The limiting molar conductivities of HCl , NaCl , and NaA are 425 S cm ^2 mol ⁻¹ , 125 S cm ^2 mol ⁻¹ , and 100 S cm ^2 mol ⁻¹ respectively. The conductivity of a 0.08 M aqueous solution of the weak acid HA is 6.4 10⁻³ S cm ⁻¹ . The acid dissociation constant K_a of HA is y 10⁻³ . The value of y is ________.

Step-by-step solution

According to Kohlrausch's law, the limiting molar conductivity of the weak acid HA is: _m^0( HA ) = _m^0( HCl ) + _m^0( NaA ) - _m^0( NaCl ) _m^0( HA ) = 425 + 100 - 125 = 400 S cm ^2 mol ⁻¹ The molar conductivity of the HA solution at concentration c = 0.08 M is: _m = 1000 c _m = 1000 6.4 10⁻³ 0.08 = 6.4 0.08 = 80 S cm ^2 mol ⁻¹ The degree of dissociation ( ) is given by: = _m _m^0 = 80 400 = 0.2 The acid dissociation constant K_a is calculated using Ostwald's dilution law: K_a = c ^2 1 - K_a = 0.08 (0.2)^2 1 - 0.